Question

Seasonal affective disorder (SAD) is a type of depression during seasons with less daylight (e.g., winter months). One therapy for SAD is phototherapy, which is increased exposure to light used to improve mood. A researcher tests this therapy by exposing a sample of patients with SAD to different intensities of light (low, medium, high) in a light box, either in the morning or at night (these are the times thought to be most effective for light therapy). All participants rated their mood following this therapy on a scale from 1 (poor mood) to 9 (improved mood). The hypothetical results are given in the following table.

		Light Intensity
		Low	Medium	High
Time of Day	Morning	5	5	7
		6	6	8
		3	4	6
		7	7	8
		5	9	5
		6	8	8
	Night	5	6	8
		8	8	7
		6	7	6
		7	5	8
		4	9	7
		3	8	6

(a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Round your answers to two decimal places. Assume experimentwise alpha equal to 0.05.)

Source of Variation	SS	df	MS	F
Time of day
Intensity
Time of day × Intensity
Error
Total

State the decision for the main effect of the time of day.

Retain the null hypothesis.Reject the null hypothesis.     

State the decision for the main effect of intensity.

Retain the null hypothesis.Reject the null hypothesis.     

State the decision for the interaction effect.

Retain the null hypothesis.Reject the null hypothesis.     

(b) Compute Tukey's HSD to analyze the significant main effect.

The critical value is  for each pairwise comparison.

Answer 1

Anova: Two-Factor With Replication
SUMMARY	Low	Medium	High	Total
Morning
Count	6	6	6	18
Sum	32	39	42	113
Average	5.333333	6.5	7	6.277778
Variance	1.866667	3.5	1.6	2.565359
Night
Count	6	6	6	18
Sum	33	43	42	118
Average	5.5	7.166667	7	6.555556
Variance	3.5	2.166667	0.8	2.496732
Total
Count	12	12	12
Sum	65	82	84
Average	5.416667	6.833333	7
Variance	2.44697	2.69697	1.090909
ANOVA
Source of Variation	SS	df	MS	F	P-value
Time of day	0.69	1	0.69	0.31	0.5817
Intensity	18.17	2	9.08	4.06	0.0276
Time of day x Intensity	0.72	2	0.36	0.16	0.8518
Error	67.17	30	2.24
Total	86.75	35

Decision for the main effect of the time of day: Retain the null hypothesis.

Decision for the main effect of intensity: Reject the null hypothesis.   

Decision for the interaction effect: Retain the null hypothesis.

(b) Compute Tukey's HSD to analyze the significant main effect.

At α = 0.05, k = 3, N-K = 30, Q value = 3.49

Critical value, CV = Q*√(MSW/n) = 3.49*√(2.24/12) = 1.51